Calculating Specific Volume of Hydrogen at Given Pressure and Temperature?

AI Thread Summary
To calculate the specific volume of hydrogen at a pressure of 1000 mb and a temperature of +10° C, the ideal gas law PV=nRT is utilized. The sample is identified as 1 gram of hydrogen, which corresponds to 0.5 moles of H2 due to its molecular weight of 2.016 g/mol. Using the values for R, T, and P, the specific volume is derived directly from the equation V=nRT/P. The discussion highlights that the specific volume can be calculated without needing to determine density first. The final result confirms that the specific volume of hydrogen is effectively calculated as 11.77163 L/g.
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Homework Statement



A sample of hydrogen is at a pressure of 1000mb and a temperature of +10° C.
Calculate its specific volume.

Homework Equations



I'm guessing PV=mRT

The Attempt at a Solution



P=1 bar
m=2 g
R=constant (do I use the universal constant or is there a constant for Hydrogen? If so how do I find it)
T=283.15K
 
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The ideal gas law is PV=nRT where n is number of gas moles. You can use the ideal gas law, but you first have to find how many moles of hydrogen are in 2kg of hydrogen. To do this, you can use this conversion factor for hydrogen:

\displaystyle \frac{1.008 g}{mole}Hydrogen
 
Ok so I don't know why I typed 2kg of hydrogen. Its a sample of hydrogen which is 1 gram but since Hydrogen is H2 in the atmosphere it should be 2 grams which is 2 moles correct?
 
If the problem states that it's a 1 gram sample of Hydrogen, then the mass is 1 gram. Hydrogen gas is indeed H2, which means each molecule contains two Hydrogen atoms. To account for this, you make a new conversion factor for H2, multiplying the mass of hydrogen by 2.

\displaystyle \frac{2.016 g}{mole}H_2
 
Ok redid some work

Density = (P * MW) / (R * T)
0.9869231693139999 atm
Molecular weight of 2 g/mol
R = 0.08206 L*ATM / mol*K
T = 283.15K

1.97384/23.235289 = 0.08495 g/L

specific volume = 1/density
specific volume of hydrogen = 11.77163 g/L

Maybe yes?
 
It looks alright, but when you take 1/density at the end, the units change to L/g. However, there was no need to use the density equation. Allow me to show you a simpler way:

\displaystyle 1g\: H_2 ×\frac{1\: mole}{2.016 g}=0.5\:mole\: H_2

\displaystyle PV=nRT

\displaystyle V=\frac{nRT}{P}

n is the number of moles of H2 (0.5), R is the gas constant, and T and P are given. This gives the volume of one gram of hydrogen gas, which is equal to the specific volume.
 
Awesome, thanks for explaining an easier way too!
 
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